Suppose we wanted to evaluate the double integral $S = \iint_D e^{x^2 + y^2} \, dx \, dy$ by first applying a change of variables from $D$ to $R$ : $\begin{aligned} x &= X_1(u, v) = u\cos(v) \\ \\ y &= X_2(u, v) = u\sin(v) \end{aligned}$ What is $S$ under the change of variables? Assume $u < 0$. If you know an expression within absolute value is non-negative, do not use absolute value at all. $S = \iint_R $ $du \, dv$
Answer: If we have a transformation $\bold{X} : R \to D$, then we can rewrite an integral under the change of variables: $ \iint_D f(x, y) \, dA = \iint_R f(\bold{X}(u, v)) | J(\bold{X}) | \, du \, dv$ First we need to find the absolute value of the Jacobian, $|J(\bold{X})|$. We'll use the Pythagorean identity, which says that $\cos^2(v) + \sin^2(v) = 1$. $\begin{aligned} |J(\bold{X})| &= \left| \det \begin{pmatrix} \dfrac{\partial X_1}{\partial u} & \dfrac{\partial X_1}{\partial v} \\ \\ \dfrac{\partial X_2}{\partial u} & \dfrac{\partial X_2}{\partial v} \end{pmatrix} \right| \\ \\ &= \left| \det \begin{pmatrix} \cos(v) & -u\sin(v) \\ \\ \sin(v) & u\cos(v) \end{pmatrix} \right| \\ \\ &= \left| u\cos^2(v) + u\sin^2(v) \right| \\ \\ &= \left| u(\cos^2(v) + \sin^2(v)) \right| \\ \\ &= \left| u \right| \end{aligned}$ We are given that $u < 0$, so $\left| u \right|$ simplifies to $-u$. Now we substitute $u$ and $v$ in $f(x, y)$. Again we use the Pythagorean identity to simplify. $\begin{aligned} f(x, y) &= f(\bold{X}(u, v)) \\ \\ &= e^{X_1(u, v)^2 + X_2(u, v)^2} \\ \\ &= e^{u^2\cos^2(v) + u^2\sin^2(v)} \\ \\ &= e^{u^2(\cos^2(v) + \sin^2(v))} \\ \\ &= e^{u^2} \end{aligned}$ Putting everything together, we get the integral under the change of variables: $ \iint_R -u e^{u^2} \, du \, dv$